NDC Mathematics Chief Solves FLT (Fermat’s Last Theorem) with Simple (Non-Abstract) Algebra

A Non–Abstract Algebraic Proof of Fermat’s Last Theorem

By DM Delinferni, Jr. Chief, Pure and Applied Mathematics Branch, National Diabetes Center (dmd@delinferni.diabetes-mellitus.org)

See also https://diabetes-mellitus.org/nonabstractalgflt.pdf

The only universal restriction to the set forming Pythagorean Triples [of the (n=2) second order] can be represented by the following equivalent identities labeled as (#1) and (#2) respectively:-

For any base x where X₀²+Y₀²=Z₀ ²and a=Z₀-Y₀  

#1) (2ax₀+a²)² + (2ax₀²+2a² x₀+(a³ – a)/2)² = (2ax₀²+2a² x₀+(a³ – a)/2)²

Consider two right-triangles, A and B which are related in that

a) the base of the larger triangle (B) is the square of the base of the smaller triangle (A)

b) the difference between the hypotenuse and the vertical arms of both triangles is the same and equal to some measure, “a”

c) we will be interested in framing the vertical arms, “x” and “y”, of the both triangles, (A) and (B), as functions of the vertical arm, “d”, of the smaller triangle, (A), as well as of the difference, “a”.

x = (=x^1/2)² = (d + a)² – d² = (2ad + a²) = 2a ∙ a²_d

x² = (y + a)² – y² = (2ay + a²) = (2ad + a²)² = 4 a²d²+ 4a³d + a⁴

2ay = 4 a²d²+ 4a³d + a⁴ – a²

y = 2ad² + 2a²d + (a³ – a)/2 = 2a ∙ 2a² ∙ (a³ – a)/2_d

z = y + a = 2a ∙ 2a² ∙ (a³ + a)/2_d

And if Z₁ – Y₁ =1, substituting k=a^-1 and x₁=x₀+(a-1)/2, then :-

#2) k[( 2x₁+1)²  + ( 2x₁²+ 2x₁)²   =( 2x₁²+ 2x₁ +1)²]

or where subscripts represent the base variable “x₁“ and “∙“ separates its powers,

        #2a) k[( 2∙1_x1)²  + ( 2∙2∙0_x1)²   = ( 2∙2∙1_x1)²]

Then, if some X₁ⁿ + Y₁ⁿ   = Z₁ⁿ  were to represent another true identity, it would of necessity map to any other true identity including, but not limited to,

X₀² + Y₀² = Z₀².  So, dividing both sides of  X₁ⁿ + Y₁ⁿ   = Z₁ⁿ  by k=(Z₁-Y₁),  

and letting A= X₁/k, B= Y₁/k, and C= Z₁/k, then  (C-B) = 1, and :-

k(Aⁿ + Bⁿ = Cⁿ)

Therefore via identity #2a

k[ (ⁿ√(2∙1_x1)²)ⁿ  + (ⁿ√ (2∙2∙0_x1)² )ⁿ  = (ⁿ√ (2∙2∙1_x1)²)ⁿ]

substituting  A=ⁿ√(2∙1_x1)²   , B= ⁿ√ (2∙2∙0_x1)²   and C= ⁿ√ (2∙2∙0_x1)²,

then in order for A, B, and C to all be non-zero and rational with any odd n>2, all three (A,B, and C) would of necessity have to have rational 2n^th roots – which seems on its face to be absurd as B and C only differ by 1.  This is confirmed via a very nice proof offered by Professor A. Rosenbloom of the University of Toronto, since B and C differ by 1, and  B=Y²ⁿ, and C=B+1=Z²ⁿ, then 1 = Z²ⁿ  – Y²ⁿ and 1=(Zⁿ+Yⁿ)( Zⁿ-Yⁿ).  Therefore both (Zⁿ+Yⁿ) and ( Zⁿ-Yⁿ) must divide  and equal 1 implying either Z or Y =0 and the other =1.

And, therefore,  FLT must hold for non-zero rationals with any odd n>2.